How to Compare Enthalpy and Entropy in Gibbs Energy

An ice cube melts in a glass of water, even though melting absorbs energy. If reactions and physical changes always favoured releasing heat, the ice should stay frozen. So why does it melt at \(5^\circ\mathrm{C}\) but freeze at \(-5^\circ\mathrm{C}\)?

The answer is that enthalpy and entropy compete. To compare their contributions to Gibbs free energy, calculate \(\Delta H\) and \(T\Delta S\) in the same units, then subtract:

\[ \Delta G=\Delta H-T\Delta S \]

If \(\Delta G<0\), the forward process is thermodynamically favourable under the specified conditions. If \(\Delta G>0\), the reverse direction is favoured. Temperature can change which contribution dominates, even when the enthalpy and entropy changes stay approximately constant.

01How to compare the two contributions

Imagine two influences acting on a chemical change.

Enthalpy tells us whether the system absorbs or releases heat at constant pressure. Entropy describes how the number of accessible microscopic arrangements of matter and energy changes.

A process can absorb heat and still be favourable if its entropy increase makes a sufficiently large contribution.

The Gibbs equation puts those effects on the same energy scale:

\[ \Delta G=\Delta H-T\Delta S \]

Here:

• \(\Delta G\) is the Gibbs free energy change, in \(\mathrm{kJ\,mol^{-1}}\).
• \(\Delta H\) is the enthalpy change, in \(\mathrm{kJ\,mol^{-1}}\).
• \(T\) is the absolute temperature, in kelvin (\(\mathrm{K}\)).
• \(\Delta S\) is the entropy change, usually supplied in \(\mathrm{J\,mol^{-1}\,K^{-1}}\).

These molar changes refer to the process or balanced reaction as written.

For a chemical reaction using standard reference data, write the relationship as:

\[ \Delta G^\circ=\Delta H^\circ-T\Delta S^\circ \]

The superscript \(\circ\) indicates standard-state values. Standard-state data describe substances in defined reference states at a specified temperature, rather than every possible reaction mixture.

The sign of each contribution

The easiest way to avoid mistakes is to look at the two terms that are actually added together: \(\Delta H\) and \(-T\Delta S\).

Because temperature in kelvin is positive, a positive \(\Delta S\) produces a negative \(-T\Delta S\).

That is the part students often reverse.

Suppose a process has:

\[ \Delta H=+20\ \mathrm{kJ\,mol^{-1}} \]

and, at the stated temperature,

\[ T\Delta S=+27\ \mathrm{kJ\,mol^{-1}} \]

The enthalpy contribution is an unfavourable \(+20\), but the entropy contribution to Gibbs energy is a favourable \(-27\).

Therefore,

\[ \Delta G=20-27=-7\ \mathrm{kJ\,mol^{-1}} \]

The entropy contribution outweighs the enthalpy cost by \(7\ \mathrm{kJ\,mol^{-1}}\).

Notice that \(-7\) is the net Gibbs energy change, not an additional quantity of heat released.

The unit conversion you must make

Entropy is often given in joules, while enthalpy is given in kilojoules.

Before calculating, convert:

\[ \Delta S_{\mathrm{kJ}} =\frac{\Delta S_{\mathrm{J}}}{1000} \]

For example,

\[ 80.0\ \mathrm{J\,mol^{-1}\,K^{-1}} \]

becomes

\[ 0.0800\ \mathrm{kJ\,mol^{-1}\,K^{-1}} \]

Multiplying by temperature then produces \(\mathrm{kJ\,mol^{-1}}\), the same unit as enthalpy.

A useful check is that Gibbs energy, enthalpy, and \(T\Delta S\) must all have matching energy units before you compare them.

NSW syllabus context (2017 edition): Gibbs free energy calculations and temperature-dependent spontaneity are explicitly included in Year 11 Module 4, Drivers of Reactions, of the Chemistry Stage 6 Syllabus (2017). That syllabus remains in use in 2026 and 2027. The Chemistry 11-12 Syllabus (2025) begins implementation in 2028, with its first HSC examination in 2029. This article teaches the 2017 syllabus method; it should not be read as a claim that the revised syllabus assesses identical content.

If the signs of \(\Delta H\) and \(\Delta S\) are still unfamiliar, Entropy and Enthalpy Changes: HSC Chemistry Explained develops those two ideas separately.

02The experiment that shows why heat alone is not enough

Return to the ice cube.

Our first prediction might be that melting is unfavourable because it requires energy to overcome some of the intermolecular attractions holding water molecules in a solid structure.

Yet ice melts when the surroundings are warm enough.

What could we measure to understand that apparent contradiction?

The first design: ice in an open glass

Place ice in a glass of water and monitor the temperature using a thermometer.

As heat flows from the warmer surroundings into the ice, melting occurs. The water temperature may decrease as it transfers energy to the ice.

This establishes something useful: melting absorbs energy.

However, an open glass does not provide a controlled comparison between temperatures. Heat enters from the room, the glass, and the surrounding air. The temperature changes during the experiment, and the amount of heat exchanged is not necessarily known.

We need to control temperature rather than just observe it changing.

The revised design: two controlled-temperature baths

Consider two identical transparent tubes, each initially containing pure ice and liquid water. The tubes are capped, and the pressure is approximately atmospheric.

One tube is placed in a temperature-controlled bath at \(278\ \mathrm{K}\), approximately \(5^\circ\mathrm{C}\).

The other is placed in a refrigerated bath at \(268\ \mathrm{K}\), approximately \(-5^\circ\mathrm{C}\).

Each bath contains a suitable circulating temperature-control fluid and is maintained at its set temperature.

Temperature probes monitor the baths, while the changing proportions of ice and liquid water are observed.

What would you expect to happen?

At \(278\ \mathrm{K}\), the ice tends to melt as heat enters from the warmer bath.

At \(268\ \mathrm{K}\), the liquid water tends to freeze as heat passes into the colder bath.

These are established behaviours of pure water near atmospheric pressure. The apparatus describes how they could be compared; the illustration and numerical results below are not measurements from a newly performed experiment.

The important finding is that melting has the same positive enthalpy change in both cases, to a close approximation. What changes is the temperature multiplying the entropy increase.

What the experiment can and cannot tell us

The controlled baths establish which direction of phase change is favoured at each temperature.

They do not, by themselves, measure the entropy change.

The enthalpy of fusion can be determined using calorimetry. At the reversible melting temperature, the solid and liquid phases can coexist in equilibrium, giving:

\[ \Delta G_{\mathrm{fus}}=0 \]

Therefore,

\[ \Delta S_{\mathrm{fus}} =\frac{\Delta H_{\mathrm{fus}}}{T_{\mathrm{m}}} \]

where \(T_{\mathrm{m}}\) is the equilibrium melting temperature, and the subscript \(\mathrm{fus}\) refers to fusion, or melting.

This lets us connect measurable heat and phase equilibrium to entropy.

The controlled-temperature experiment tests the predicted change in direction. It is not an independent measurement of every quantity in the Gibbs equation.

04How temperature changes which contribution dominates

The Gibbs relationship is especially useful when enthalpy and entropy favour opposite directions.

For a positive entropy change, increasing temperature makes \(-T\Delta S\) more negative. The entropy contribution becomes increasingly favourable.

For a negative entropy change, increasing temperature makes \(-T\Delta S\) more positive. The entropy contribution becomes increasingly unfavourable.

This gives four possibilities.

These predictions assume the stated enthalpy and entropy changes remain approximately constant. They also refer to the reaction or process under the specified thermodynamic conditions.

Real substances can undergo additional phase changes, and thermodynamic quantities may vary with temperature. A threshold calculated from constant values is therefore an estimate unless those assumptions are justified.

Finding the crossover temperature

When both contributions have opposite effects on Gibbs energy, the temperature at which they balance is found by setting \(\Delta G=0\).

\[ 0=\Delta H-T\Delta S \]

Rearrange:

\[ T_{\mathrm{cross}} =\frac{\Delta H}{\Delta S} \]

Both values must be in compatible units.

A physically meaningful positive crossover temperature requires \(\Delta H\) and \(\Delta S\) to have the same sign.

The crossover does not automatically mean that an experimental sample will transform instantly at that temperature. It identifies a thermodynamic boundary for the specified process or standard-state comparison.

06Gibbs energy tells you feasibility, not reaction speed

Imagine two processes with negative Gibbs energy changes.

One happens almost immediately. The other barely progresses during the time you observe it.

There is no contradiction.

Gibbs energy compares the thermodynamic tendency of the initial and final states. Reaction rate describes how quickly the system moves between them.

A reaction may require particles to overcome a substantial activation-energy barrier even if the products have lower Gibbs energy.

A catalyst can provide a faster reaction pathway by lowering the activation-energy barrier. It does not change the Gibbs energy difference between reactants and products, or the equilibrium constant at a fixed temperature.

There is another important distinction.

Standard Gibbs energy is not the Gibbs energy of every mixture

For reactions, the commonly supplied equation is:

\[ \Delta G^\circ=\Delta H^\circ-T\Delta S^\circ \]

A negative \(\Delta G^\circ\) means the forward reaction is thermodynamically favourable for the specified standard-state comparison.

The Gibbs energy change in an actual reacting mixture also depends on composition, such as the concentrations or partial pressures of the reactants and products.

Therefore, a positive \(\Delta G^\circ\) does not mean that no product can ever form.

Similarly, a negative \(\Delta G^\circ\) does not guarantee complete conversion into products.

For an actual system at equilibrium, the Gibbs energy change for an infinitesimal forward reaction is zero. Forward and reverse reactions can still continue at equal rates.

The ice-melting calculation is simpler because it compares two pure phases at the same temperature and pressure. There is no changing reaction-mixture concentration to include.

A useful decision process

When solving a Gibbs energy question:

1. Write the forward process. Decide which change the supplied \(\Delta H\), \(\Delta S\), and \(\Delta G\) refer to.
2. Convert the units. Use kelvin for temperature, and ensure \(\Delta H\) and \(T\Delta S\) are both in \(\mathrm{kJ\,mol^{-1}}\).
3. Calculate the signed entropy contribution. A negative \(\Delta S\) makes \(-T\Delta S\) positive.
4. Subtract and interpret. Calculate \(\Delta G=\Delta H-T\Delta S\), identify which contribution dominates, and state the thermodynamic conclusion.
5. Check the conditions. Decide whether the question concerns standard states, an actual reaction mixture, or a phase change. Keep feasibility separate from rate.

If your enthalpy data come from bond calculations, remember that average bond enthalpies are estimates. Bond Enthalpy and Conservation of Energy in HSC Chemistry explains where those estimates come from and why physical states matter.

07Questions and solutions

The questions become progressively more demanding. All numerical data identified as illustrative are invented for these questions, not experimental measurements or standard reference values.

Question 1

A hypothetical ionic compound dissolves in water:

\[ \ce{MX(s) -> M+(aq) + X-(aq)} \]

At \(300\ \mathrm{K}\), the illustrative standard thermodynamic data are:

\[ \Delta H^\circ=+18.0\ \mathrm{kJ\,mol^{-1}} \]

\[ \Delta S^\circ=+80.0\ \mathrm{J\,mol^{-1}\,K^{-1}} \]

Calculate \(\Delta G^\circ\), identify the dominant contribution, and explain whether dissolution is thermodynamically favourable in the standard-state comparison.

Solution 1

The standard Gibbs energy change is \(-6.0\ \mathrm{kJ\,mol^{-1}}\). The favourable entropy contribution outweighs the endothermic enthalpy cost.

First convert the entropy change:

\[ \Delta S^\circ =+0.0800\ \mathrm{kJ\,mol^{-1}\,K^{-1}} \]

Then calculate:

\[ \begin{aligned} \Delta G^\circ &=\Delta H^\circ-T\Delta S^\circ\\ &=18.0-300(0.0800)\\ &=18.0-24.0\\ &=-6.0\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

The enthalpy contribution is \(+18.0\), which opposes dissolution.

The entropy contribution to Gibbs energy is \(-24.0\), which favours dissolution.

Because the entropy benefit is larger, the standard Gibbs energy change is negative.

The tempting mistake is to conclude that an endothermic dissolution must be unfavourable. That ignores the positive entropy change and its increasing importance at higher temperature.

The result indicates thermodynamic favourability for the specified standard-state comparison, not how quickly the solid dissolves or how much can dissolve in a particular solution.

Question 2

A student investigates the dissolution of a hypothetical soluble ionic compound in a simple insulated calorimeter.

A \(0.0500\ \mathrm{mol}\) sample dissolves in water. The final solution has a mass of \(100\ \mathrm{g}\).

The temperature falls from \(298\ \mathrm{K}\) to \(295\ \mathrm{K}\).

Assume:

• The calorimeter is perfectly insulated, and its own heat capacity is negligible.
• The solution’s specific heat capacity is \(4.18\ \mathrm{J\,g^{-1}\,K^{-1}}\).
• The observed molar enthalpy of dissolution approximates \(\Delta H^\circ\).
• The illustrative standard entropy change is \(+100\ \mathrm{J\,mol^{-1}\,K^{-1}}\).
• Enthalpy and entropy changes are approximately constant over the temperature range.

(a) Calculate the molar enthalpy of dissolution.

(b) Estimate \(\Delta G^\circ\) at the initial temperature of \(298\ \mathrm{K}\), and identify which contribution dominates.

(c) In a second trial, an open, poorly insulated glass beaker produces a temperature fall of only \(1.8\ \mathrm{K}\) for the same amount of compound and solution. Explain why the second trial may underestimate the enthalpy absorbed.

(d) Explain why neither temperature measurement alone establishes the entropy change or the time needed for complete dissolution.

Solution 2

The estimated molar enthalpy of dissolution is \(+25.1\ \mathrm{kJ\,mol^{-1}}\), and \(\Delta G^\circ\) at \(298\ \mathrm{K}\) is approximately \(-4.72\ \mathrm{kJ\,mol^{-1}}\). Entropy outweighs the heat-absorbing cost.

(a) Enthalpy from calorimetry

The heat change of the solution is:

\[ q_{\mathrm{sol}}=mc\Delta T \]

Here, \(q_{\mathrm{sol}}\) is the heat gained by the solution, \(m\) its mass, and \(c\) its specific heat capacity.

The temperature change is:

\[ \Delta T=295-298=-3\ \mathrm{K} \]

Therefore,

\[ \begin{aligned} q_{\mathrm{sol}} &=100(4.18)(-3)\\ &=-1254\ \mathrm{J}\\ &=-1.254\ \mathrm{kJ} \end{aligned} \]

The solution loses heat, so the dissolving process absorbs the same amount under the insulation assumptions:

\[ q_{\mathrm{diss}}=+1.254\ \mathrm{kJ} \]

Divide by the amount dissolved:

\[ \begin{aligned} \Delta H_{\mathrm{diss}} &=\frac{1.254}{0.0500}\\ &=+25.08\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

The estimated enthalpy of dissolution is approximately \(+25.1\ \mathrm{kJ\,mol^{-1}}\).

(b) Gibbs energy at \(298\ \mathrm{K}\)

Convert entropy:

\[ \Delta S^\circ =+0.100\ \mathrm{kJ\,mol^{-1}\,K^{-1}} \]

Use the estimated enthalpy:

\[ \begin{aligned} \Delta G^\circ &\approx25.08-298(0.100)\\ &=25.08-29.80\\ &=-4.72\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

The entropy contribution of \(-29.80\ \mathrm{kJ\,mol^{-1}}\) outweighs the enthalpy cost of \(+25.08\ \mathrm{kJ\,mol^{-1}}\).

The process is therefore favourable in the approximate standard-state comparison at \(298\ \mathrm{K}\).

(c) Why the open beaker underestimates enthalpy

The dissolving process cools the solution below room temperature.

In an open beaker, heat can enter from the warmer surroundings during dissolution. That additional heat reduces the observed temperature drop.

If the student assumes all heat absorbed came from the solution, the calculated heat absorption will be too small.

The reported molar enthalpy will consequently be less positive than the true value.

(d) What cannot be concluded from the thermometer

The thermometer records a temperature change. It does not directly measure the entropy change or the number of accessible microscopic arrangements.

An entropy value requires additional thermodynamic information, such as suitable standard molar entropy data.

Also, a negative Gibbs energy change is not a rate measurement. Dissolution speed depends on factors such as particle size, mixing, diffusion, and conditions at the solid surface.

The tempting conclusion is that a cooling solution must be thermodynamically unfavourable or slow. Neither follows from the observed temperature drop alone.

Question 3

A hypothetical metal carbonate undergoes thermal decomposition:

\[ \ce{MCO3(s) -> MO(s) + CO2(g)} \]

The illustrative standard thermodynamic data are:

\[ \Delta H^\circ=+160\ \mathrm{kJ\,mol^{-1}} \]

\[ \Delta S^\circ=+170\ \mathrm{J\,mol^{-1}\,K^{-1}} \]

A small furnace maintains a sample at \(900\ \mathrm{K}\). A pump continuously removes carbon dioxide, and gas production is observed.

A student argues:

“The reaction releases carbon dioxide, so its Gibbs energy change must be negative at \(900\ \mathrm{K}\).”

(a) Calculate \(\Delta G^\circ\) at \(900\ \mathrm{K}\) and \(1000\ \mathrm{K}\).

(b) Explain why gas production at \(900\ \mathrm{K}\) does not necessarily contradict the calculated standard Gibbs energy.

(c) Would adding a catalyst change the crossover temperature calculated from the supplied standard thermodynamic data? Explain.

Assume \(\Delta H^\circ\) and \(\Delta S^\circ\) are approximately constant over this temperature range.

Solution 3

The standard Gibbs energy is positive at \(900\ \mathrm{K}\) and negative at \(1000\ \mathrm{K}\). Carbon dioxide removal can nevertheless allow the reaction to proceed at \(900\ \mathrm{K}\), because the actual gas pressure differs from the standard state.

(a) Gibbs energy at both temperatures

Convert entropy:

\[ \Delta S^\circ =+0.170\ \mathrm{kJ\,mol^{-1}\,K^{-1}} \]

At \(900\ \mathrm{K}\):

\[ \begin{aligned} \Delta G^\circ_{900} &=160-900(0.170)\\ &=160-153\\ &=+7.0\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

At \(1000\ \mathrm{K}\):

\[ \begin{aligned} \Delta G^\circ_{1000} &=160-1000(0.170)\\ &=160-170\\ &=-10.0\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

The positive entropy change makes decomposition increasingly favourable as temperature increases.

(b) Why gas production is possible at \(900\ \mathrm{K}\)

The student has confused standard Gibbs energy with Gibbs energy under the actual experimental conditions.

For this reaction, the carbon dioxide partial pressure affects the thermodynamic driving force.

The standard Gibbs energy calculation assumes the specified standard reference pressure for the gas. The pump instead maintains a lower carbon dioxide partial pressure.

Removing a gaseous product favours further decomposition. If the pressure is sufficiently low, the actual forward reaction can be favourable even though \(\Delta G^\circ\) is positive.

This is consistent with the equilibrium principles used elsewhere in HSC Chemistry.

Observing gas production does not, by itself, prove that \(\Delta G^\circ<0\).

(c) Effect of a catalyst

A catalyst does not change the crossover temperature predicted by the supplied thermodynamic data.

The calculated crossover is:

\[ \begin{aligned} T_{\mathrm{cross}} &=\frac{160}{0.170}\\ &\approx941\ \mathrm{K} \end{aligned} \]

A catalyst affects the activation-energy barrier and therefore the reaction rate.

It does not alter the enthalpy and entropy differences between the specified initial and final states.

The tempting explanation is that the catalyst somehow makes the reaction’s Gibbs energy more negative. It does not.

The observed decomposition at \(900\ \mathrm{K}\) is instead explained by the non-standard carbon dioxide pressure, while the catalyst, if used, could affect how quickly the system responds.

Question 4

A hypothetical endothermic process takes place in a closed reaction vessel held at constant temperature and pressure by contact with a large thermal bath.

Its illustrative thermodynamic changes are:

\[ \Delta H=+42.0\ \mathrm{kJ\,mol^{-1}} \]

\[ \Delta S_{\mathrm{sys}} =+120\ \mathrm{J\,mol^{-1}\,K^{-1}} \]

Here, \(\Delta S_{\mathrm{sys}}\) is the entropy change of the reacting system.

For an ideal thermal bath at temperature \(T\), assume:

\[ \Delta S_{\mathrm{surr}} =-\frac{\Delta H}{T} \]

where \(\Delta S_{\mathrm{surr}}\) is the entropy change of the surroundings. The total entropy change is the sum of the system and surroundings contributions.

(a) At \(300\ \mathrm{K}\), calculate the entropy change of the surroundings, total entropy change, and Gibbs energy change.

(b) Repeat at \(400\ \mathrm{K}\).

(c) A student claims the process must be favourable at both temperatures because the system’s entropy increases in both cases. Explain the flaw.

(d) Would the \(400\ \mathrm{K}\) prediction necessarily remain valid if the reacting system were thermally isolated instead of being held at constant temperature? Explain.

Solution 4

The process is unfavourable at \(300\ \mathrm{K}\) but favourable at \(400\ \mathrm{K}\). The system’s entropy increases at both temperatures, but the entropy change of the surroundings changes the total result.

(a) At \(300\ \mathrm{K}\)

The reaction absorbs \(42\,000\ \mathrm{J}\) per mole of reaction as written.

For an ideal bath, the surroundings lose that heat.

\[ \begin{aligned} \Delta S_{\mathrm{surr}} &=-\frac{42000}{300}\\ &=-140\ \mathrm{J\,mol^{-1}\,K^{-1}} \end{aligned} \]

The total entropy change is:

\[ \begin{aligned} \Delta S_{\mathrm{total}} &=120-140\\ &=-20\ \mathrm{J\,mol^{-1}\,K^{-1}} \end{aligned} \]

The negative total entropy change indicates that the forward process is unfavourable under these fixed-temperature conditions.

Calculate Gibbs energy independently:

\[ \begin{aligned} \Delta G_{300} &=42.0-300(0.120)\\ &=42.0-36.0\\ &=+6.0\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

The positive Gibbs energy confirms the conclusion.

(b) At \(400\ \mathrm{K}\)

The surroundings still lose \(42\,000\ \mathrm{J}\) per mole of reaction, but their entropy decreases by a smaller amount because the temperature is higher.

\[ \begin{aligned} \Delta S_{\mathrm{surr}} &=-\frac{42000}{400}\\ &=-105\ \mathrm{J\,mol^{-1}\,K^{-1}} \end{aligned} \]

Now:

\[ \begin{aligned} \Delta S_{\mathrm{total}} &=120-105\\ &=+15\ \mathrm{J\,mol^{-1}\,K^{-1}} \end{aligned} \]

The total entropy increases, so the forward process is favourable.

Using Gibbs energy:

\[ \begin{aligned} \Delta G_{400} &=42.0-400(0.120)\\ &=42.0-48.0\\ &=-6.0\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

The negative Gibbs energy agrees with the positive total entropy change.

(c) Why system entropy alone is insufficient

The student considers only \(\Delta S_{\mathrm{sys}}\).

At \(300\ \mathrm{K}\), the system gains \(120\ \mathrm{J\,mol^{-1}\,K^{-1}}\) of entropy, but the surroundings lose \(140\ \mathrm{J\,mol^{-1}\,K^{-1}}\).

That loss is larger than the system’s gain.

At \(400\ \mathrm{K}\), the same heat removal causes a smaller entropy decrease in the surroundings. The system’s entropy gain is then larger than the surroundings’ entropy loss.

The deeper connection is:

\[ \Delta G=-T\Delta S_{\mathrm{total}} \]

for the fixed-temperature, fixed-pressure conditions assumed here, with the surroundings treated as an ideal thermal reservoir.

This shows why Gibbs energy combines enthalpy and entropy in precisely the way it does.

(d) What changes in an isolated system

The \(400\ \mathrm{K}\) prediction cannot automatically be applied throughout an isolated process.

The reaction is endothermic, so it requires an energy source.

Without a thermal bath supplying heat, the reacting system may cool as the reaction proceeds. Temperature would no longer remain fixed at \(400\ \mathrm{K}\).

Because the entropy contribution depends on temperature, the Gibbs energy comparison would also change.

The tempting mistake is to use a fixed-temperature equation while quietly allowing the temperature to change.

The heat bath is not incidental equipment. It is part of the condition that makes the thermodynamic comparison valid.

Question 5

A hypothetical pure substance X can exist as a solid, liquid, or gas at a fixed pressure.

Assume the following illustrative thermodynamic data remain constant over the temperature range considered.

The substance is maintained at constant pressure and temperature, and the three possible phases are considered as competing thermodynamic states. Ignore supercooling, superheating, and kinetic barriers when identifying the stable equilibrium phase.

A student combines the two transformations to calculate the Gibbs energy change for:

\[ \ce{X(s) -> X(g)} \]

They find that solid and gas have equal Gibbs energies at approximately \(379\ \mathrm{K}\).

The student concludes:

“At \(379\ \mathrm{K}\), solid and gas must coexist as the stable equilibrium phases. The substance can change directly from solid to gas without becoming liquid.”

(a) Calculate the Gibbs energy changes for melting, vaporisation, and direct solid-to-gas conversion at \(379\ \mathrm{K}\).

(b) Determine whether the student’s conclusion is correct. Identify the thermodynamically most stable phase.

(c) Determine the stable phase at \(250\ \mathrm{K}\), \(350\ \mathrm{K}\), and \(450\ \mathrm{K}\).

(d) Find the two actual equilibrium transition temperatures predicted by the model.

(e) Explain why the direct solid-gas Gibbs energy crossover does not represent a stable phase transition at this pressure. If gas is cooled slowly from \(450\ \mathrm{K}\), what equilibrium sequence of phases is predicted?

Solution 5

The student’s conclusion is incorrect. At approximately \(379\ \mathrm{K}\), solid and gas have almost equal Gibbs energies, but liquid has a lower Gibbs energy than either. The liquid is therefore the stable phase.

The key is that equality between two possible states does not establish stable coexistence when a third state is thermodynamically preferred.

(a) Calculate the three Gibbs energy changes

For melting:

\[ \ce{X(s) -> X(l)} \]

Convert entropy:

\[ \Delta S_{\mathrm{melt}} =0.020\ \mathrm{kJ\,mol^{-1}\,K^{-1}} \]

Then:

\[ \begin{aligned} \Delta G_{\mathrm{melt}} &=6.0-379(0.020)\\ &=6.0-7.58\\ &=-1.58\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

Melting is favourable.

For vaporisation:

\[ \ce{X(l) -> X(g)} \]

\[ \begin{aligned} \Delta G_{\mathrm{vap}} &=30.0-379(0.075)\\ &=30.0-28.425\\ &=+1.575\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

Vaporisation is unfavourable. Liquid is favoured over gas.

Now consider direct solid-to-gas conversion.

Its overall enthalpy and entropy changes are the sums of the two stages because both are state functions:

\[ \begin{aligned} \Delta H_{\mathrm{sub}} &=6.0+30.0\\ &=36.0\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

\[ \begin{aligned} \Delta S_{\mathrm{sub}} &=20+75\\ &=95\ \mathrm{J\,mol^{-1}\,K^{-1}} \end{aligned} \]

Therefore,

\[ \begin{aligned} \Delta G_{\mathrm{sub}} &=36.0-379(0.095)\\ &=-0.005\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

This is extremely close to zero, as expected from the approximate crossover temperature.

(b) Determine the most stable phase

Choose solid as the reference state.

At \(379\ \mathrm{K}\):

• Liquid has a Gibbs energy \(1.58\ \mathrm{kJ\,mol^{-1}}\) lower than solid.
• Gas has a Gibbs energy approximately \(0.005\ \mathrm{kJ\,mol^{-1}}\) lower than solid.
• Therefore, liquid is approximately \(1.575\ \mathrm{kJ\,mol^{-1}}\) lower than gas.

Liquid has the lowest Gibbs energy.

The student’s conclusion is wrong because they compared solid and gas while ignoring the third candidate state.

(c) Determine the stable phase at each temperature

At \(250\ \mathrm{K}\):

\[ \begin{aligned} \Delta G_{\mathrm{melt}} &=6.0-250(0.020)\\ &=+1.0\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

\[ \begin{aligned} \Delta G_{\mathrm{vap}} &=30.0-250(0.075)\\ &=+11.25\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

Melting is unfavourable, and vaporisation is also unfavourable. Solid has a lower Gibbs energy than liquid, which in turn has a lower Gibbs energy than gas.

Solid is the stable phase.

At \(350\ \mathrm{K}\):

\[ \begin{aligned} \Delta G_{\mathrm{melt}} &=6.0-350(0.020)\\ &=-1.0\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

\[ \begin{aligned} \Delta G_{\mathrm{vap}} &=30.0-350(0.075)\\ &=+3.75\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

Liquid is favoured over solid, but gas is unfavourable relative to liquid.

Liquid is the stable phase.

At \(450\ \mathrm{K}\):

\[ \begin{aligned} \Delta G_{\mathrm{melt}} &=6.0-450(0.020)\\ &=-3.0\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

\[ \begin{aligned} \Delta G_{\mathrm{vap}} &=30.0-450(0.075)\\ &=-3.75\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

Both melting and vaporisation lower Gibbs energy.

Gas is the stable phase.

The predicted pattern is:

(d) Calculate the stable phase-transition temperatures

Melting reaches equilibrium when:

\[ \Delta G_{\mathrm{melt}}=0 \]

\[ \begin{aligned} T_{\mathrm{melt}} &=\frac{6.0}{0.020}\\ &=300\ \mathrm{K} \end{aligned} \]

Vaporisation reaches equilibrium when:

\[ \Delta G_{\mathrm{vap}}=0 \]

\[ \begin{aligned} T_{\mathrm{boil}} &=\frac{30.0}{0.075}\\ &=400\ \mathrm{K} \end{aligned} \]

These are the two stable phase-transition temperatures predicted by the model.

(e) Why the solid-gas crossover is hidden

For the direct solid-to-gas transformation:

\[ T_{\mathrm{sub}} =\frac{\Delta H_{\mathrm{sub}}}{\Delta S_{\mathrm{sub}}} \]

\[ \begin{aligned} T_{\mathrm{sub}} &=\frac{36.0}{0.095}\\ &\approx379\ \mathrm{K} \end{aligned} \]

This calculation is mathematically valid.

But it only identifies where solid and gas have equal Gibbs energies.

At that temperature, liquid is more stable than both.

Consequently, the solid-gas equality is not an observable stable phase boundary under these assumed conditions.

As gas cools slowly from \(450\ \mathrm{K}\), it first reaches the gas-liquid equilibrium at \(400\ \mathrm{K}\).

Below \(400\ \mathrm{K}\), liquid is the stable phase, so condensation is favoured.

The substance remains liquid as it cools towards \(300\ \mathrm{K}\).

At \(300\ \mathrm{K}\), the liquid and solid phases have equal Gibbs energies.

Below \(300\ \mathrm{K}\), solid becomes the stable phase, so freezing is favoured.

The equilibrium sequence is therefore:

\[ \ce{X(g) -> X(l) -> X(s)} \]

The surprising part is that the direct solid-gas Gibbs energy equality near \(379\ \mathrm{K}\) is real within the model, yet it does not determine the phase that actually minimises Gibbs energy.

This is the limitation of comparing only two states when more than two are available.

07What Gibbs energy helps you decide next

When you compare enthalpy and entropy contributions, the essential question is not simply whether heat is released or whether entropy increases.

It is whether the signed sum

\[ \Delta G=\Delta H-T\Delta S \]

is positive, negative, or zero under the stated conditions.

For a well-defined process, a negative Gibbs energy change identifies the thermodynamically favourable direction. But that direction can depend on temperature, pressure, composition, and which alternative states are available.

The next useful connection is chemical equilibrium. Gibbs energy explains the thermodynamic driving force, while equilibrium describes the composition reached when there is no net driving force for further change.

Keeping that distinction clear also prevents one of the most persistent mistakes in chemistry: assuming that a thermodynamically favourable reaction must be fast or must go to completion.